If $H$ is a positive, self-adjoint operator on a finite dimensional Hilbert space (i.e. a euclidean space), then $P_t\colon=e^{-tH}$, $t\geq0$, is a continuous contraction semigroup. If $H$ is a self-adjoint operator on a finite dimensional Hilbert space, then $U_t\colon=e^{-itH}$, $t\in\R$, is a continuous group of isometries. 2. Prove that
$$
\lim_{T\to\infty}\frac1T\int_0^T P_tx\,dt=
\lim_{T\to\infty}\frac1T\int_0^T U_tx\,dt
=\Prn_{\ker H}x~.
$$
Suppose $0\leq\l_1\leq\cdots\leq\l_n$ are the eigen values of $H$ with orthonormal eigen vectors $x_1,\ldots,x_n$, then for all $x\in E$:
$$
P_tx=\sum e^{-\l_jt}\la x,x_j\ra x_j
\quad\mbox{and}\quad
U_tx=\sum e^{-i\l_jt}\la x,x_j\ra x_j~.
$$
Since $e^{-\l_jt}\leq1$ and $|e^{-i\l_jt}|=1$, $P_t$ is a self-adjoint contraction and $U_t$ is an isometry. We obviously have for all $P_sP_t=P_{s+t}$ and $U_sU_t=U_{s+t}$.
2. Suppose $x\in(\ker H)^\perp$, then
$$
\frac1T\int_0^T P_tx\,dt
=\sum_{\l_j > 0}\frac1T\int_0^Te^{-\l_jt}\la x,x_j\ra x_j\,dt
=\sum_{\l_j > 0}\frac{1-e^{-\l_jT}}{\l_jT}\la x,x_j\ra x_j
=0
$$
and for $x\in\ker H$:
$$
\frac1T\int_0^T P_tx\,dt
=\sum_{\l_j = 0}\la x,x_j\ra x_j
=x~.
$$